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I have a motherboard which stores BIOS settings in a DS12887 chip with internal battery. The battery is long dead, and the pin of chip's socket is not connected to any power supply so the chip doesn't store any settings which include hard disk configuration.

I disconnected the internal battery, glued cell holder on top of the chip and connected it to battery pins, however it turns out I don't have any CR2032 batteries on hand. I want to test the computer and don't care if the settings are lost when I turn the computer off (for now). Is it safe to connect 5V directly to the chip's battery pin or should I make a quick 3V regulator?

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    It is very unlikely that the internal battery is required for the computer to operate (such computers do exist, but they're not IBM-compatibles), so if this is just for testing purposes until you can get the appropriate battery, just leave the battery out. You'll lose the settings, but you said you don't care about that. – Cody Gray Jun 15 '17 at 12:10
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    As @Colin__s correctly says, Vbat must be 2.5..4.0 V. For a temporary lash-up, you could put 5 V through two 1N4148 or 1N4001 diodes in series to get a 1.2..1.6 V drop and connect that to the DS12887 supply. – TonyM Jun 15 '17 at 13:06
  • @CodyGray, there are IBM-compatibles that will refuse to boot without the CMOS battery. – Mark Jun 15 '17 at 18:52
  • Seems that would make them not IBM-compatible, @Mark. :-) I can't say I've ever seen any of the beasts you describe. Perhaps that would make an interesting question in its own right? – Cody Gray Jun 16 '17 at 12:57
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    The internal battery does not get charged, so you could also have tested it with a couple of AA or AAA batteries too. The batteries don't have to supply much current, so if you don't have a battery holder handy you could always grab a remote control and run a couple of wires from its batteries across to the module :-) – Malvineous Jul 15 '17 at 11:04
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The datasheet for that part can be found here, it lists the maximum for VBAT as 4.0 V, so yes you do need to regulate down from 5 V.

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It's not clear from your question you're trying to do. A DS12887 "chip" has its own internal battery. It's actually not a chip, but a module containing a DS12885 chip, a crystal and a small lithium battery. The module only has one power supply connection, a +5V VCC pin and this pin must be connected to the +5V power supply during normal operation. It has no means of connecting an external battery. It does not use a CR2032 battery.

If you're saying that your motherboard has a CR2032 socket and you want connect +5V to it instead of replacing the battery with fresh a CR2032 then you shouldn't need to do anything. Like Cody Gray said, just remove the dead CR2032 battery and your motherboard should boot.

If you're saying you've broken open a DS12887 module and are attempting to replace the internal battery then you probably don't need to do anything except remove the battery for the motherboard to boot. Note that this battery is not a CR2032, it's a much smaller battery like a CR1232. The DS12885 chip inside the module just needs to be powered with +5V through its VCC pin, and this is required for it to be fully functional. The motherboard must already be supplying +5V to the DS12887 otherwise it would have never have worked.

You can confirm all this with the DS12887's datasheet. For example for the VBAT connection it says:

Connection for a Primary Battery. (DS12885 Only.) Battery voltage must be held between the minimum and maximum limits for proper operation. If a backup supply is not supplied, VBAT must be grounded. Connect the battery directly to the VBAT pin. Diodes in series between the VBAT pin and the battery may prevent proper operation.

Notice that only the DS12885 has a VBAT pin, and that it doesn't actually need to be connected to a battery. If you've cracked open a DS12887 you might need to ground the DS12885 chip's VBAT pin, but you don't need to supply it any power. Supplying +5V to battery terminals inside the module risks damaging it, supplying it a dropped down 2.5V - 4V from the power supply won't accomplish anything.

The description for the VCC pin reads:

When VCC is applied within normal limits, the device is fully accessible and data can be written and read. When VCC is below VPF reads and writes are inhibited.

Notice how the chip cannot be accessed unless its being supplied +5V through the VCC pin. This means when the computer is powered on the chip must be supplied with +5V for it to work, so this must already be the case.

If you're not sure whether you actually have DS12887 module then according to App Note 503: Replacing the DS1287/DS12887 Real-Time Clock in a Personal Computer here's how you can tell:

If your computer contains a Nickel Cadmium or a coin cell battery attached to the motherboard, or uses a battery connected to the motherboard via a wire harness, then your computer does not use a clock module. If your PC does use a Dallas clock module, it will look something like the following:

DS1287 and DS12887 clock modules

Finally, if you are in fact ultimately trying to replace the internal battery in a DS12887 then you should consider simply replacing the entire module instead. They're still being made and can be bought for $10 or less online.

  • This is a great answer. :) – Colin Jun 15 '17 at 20:45
  • I am pretty sure I clearly stated what I did to the module. In case you want to see how the result looks like, see two last pictures in this album: imgur.com/a/8ufkb Turns out I missed the table where safe range for Vbat was stated. I chose this method, because shipping new DS12887 costs 20 times more than a new CR2032 battery and takes about 30 days more than the time it took me to get the computer running. – Algimantas Jun 18 '17 at 17:13
  • @Algimantas No, you never said you cracked open the DS12887. Your use of "instead of CR2032 battery" in the subject made it sound like that it was a typical CR2032 internal battery of a PC that you were trying to replace. – Ross Ridge Jun 18 '17 at 17:32
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    Was "I disconnected the internal battery <...>" not clear enough? – Algimantas Jun 18 '17 at 17:45
  • @Algimantas No, as I said, it sounded like you were saying that you disconnected the CR2032 internal battery of your PC. Every time you say "internal battery" in your post it can be read as "CR2032 internal battery on the motherboard". It didn't help that you referred to the DS12887 as a chip instead of a module, making it look you might have misread the number on its case. – Ross Ridge Jun 18 '17 at 17:58

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